I am stuck here. It looks like some sort of a binomial coefficient but I just can't figure out how wolfram alpha produces this exact solution:
$$\sum_{k=1}^\infty \frac{n}{(n+k)(n+k+1)}=\frac{n}{n+1}$$
I am stuck here. It looks like some sort of a binomial coefficient but I just can't figure out how wolfram alpha produces this exact solution:
$$\sum_{k=1}^\infty \frac{n}{(n+k)(n+k+1)}=\frac{n}{n+1}$$
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By the telescoping sum: $$\sum_{k=1}^{+\infty} \frac{n}{(n+k)(n+k+1)}=n\sum_{k=1}^{+\infty}\left(\frac{1}{n+k}-\frac{1}{n+k+1}\right)=$$ $$=n\lim_{k\rightarrow+\infty}\left(\frac{1}{n+1}-\frac{1}{n+k+1}\right)=\frac{n}{n+1}$$